# Let $e$ be an edge of minimum weight in the connected weighted graph $G$. Every minimum spanning tree of $G$ contains $e$

Let $$e$$ be an edge of minimum weight in the connected weighted graph $$G$$. Every minimum spanning tree of $$G$$ contains $$e$$.

I have been told that this is not true. But I also know this property : Must a minimum weight spanning tree for a graph contain the least weight edge of every vertex of the graph?

Is it not true because that $$e$$ is not necessary the only minimum weight edge?

If $$G$$ contains a cycle of edges each of the same minimum weight, then it is not possible for $$G$$ to have a spanning tree without omitting at least one of those edges.
Otherwise the answer is yes: Let $$T$$ be a spanning tree of $$G$$ NOT containing $$e$$. Then $$T+e$$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $$T'$$ of lower weight.